NEET Physics Solutions For Class 11 Chapter 2 Circular Motion

Circular Motion

Fundamental parameter of circular motion

Radius Vector: The vector joining the center of the circle and the center of the particle performing circular motion is called the radius vector.

It has constant magnitude and variable direction

Angular Displacement (δθ or θ)

The angle described by the radius vector is called angular displacement.

Infinitesimal angular displacement is a vector quantity. However, finite angular displacement is a scalar quantity.

S.I Unit Radian

Dimension: M⁰L⁰T⁰

1 radian = \(\frac{360}{2 \pi}\)

No. Of revolution = \(\frac{\text { angular displacement }}{2 \pi}\)

In 1 revolution Δθ = 360º = 2π radian

In N revolution Δθ = 360º × N = 2πN radian

Clockwise rotation is taken as a negative

Anticlockwise rotation is taken as a positive

Question 1. If a particle completes one and a half revolutions along the circumference of a circle then its angular displacement is –

1. 0
2. π
3. 2π
4. 2π

Answer: = 3π

Angular Velocity (ω):

• The rate of change of angular displacement with time is called angular velocity. It is a vector quantity.
• The angle traced per unit time by the radius vector is called angular speed.

Instantaneous angular velocity = \(=\omega=\lim _{\delta t \rightarrow 0} \frac{\delta \theta}{\delta \mathrm{t}} \text { or } \omega=\frac{\mathrm{d} \theta}{\mathrm{dt}}\)

Average angular velocity

= \(\bar{\omega}=\frac{\theta_2-\theta_1}{\mathrm{t}_2-\mathrm{t}_1}=\frac{\Delta \theta}{\Delta \mathrm{t}}\)

S.I. Unit: rad/sec

Angular Velocity Dimension: M⁰L⁰T^-1

Angular Velocity Direction: Infinitesimal angular displacement, angular velocity, and angular acceleration are vector quantities whose direction is given by the right-hand rule.

Right-hand Rule: Imagine the axis of rotation to be held in the right hand with fingers curled around the axis and the thumb stretched along the axis. If the curled fingers denote the sense of rotation, then the thumb denotes the direction of the angular velocity (or angular acceleration of infinitesimal angular displacement.

Angular Acceleration (a):

The rate of change of angular velocity with time is called angular acceleration. Average angular acceleration

⇒ \(\bar{\alpha}=\frac{\omega_2-\omega_1}{t_2-t_1}=\frac{\Delta \omega}{\Delta t}\)

Instantaneous angular acceleration

⇒ \(\alpha=\frac{d \omega}{d t}=\frac{d^2 \theta}{d t^2}\)

It is a vector quantity, whose direction is along the change in direction of angular velocity.

S.I. Unit: radian/sec²

Dimension: M⁰L⁰T^-2

Relation Between Angular Velocity And Linear Velocity:

Suppose the particle moves along a circular path from point A to point B in infinitesimally small time δt. As, δt → 0, δθ → 0

∴ arc AB = chord AB i.e. displacement of the particle is along a straight line.

∴ Linear velocity, v = \(v=\lim _{\delta t\rightarrow 0} \frac{\delta s}{\delta t}\)

But, δs = r.δθ

∴ v = \(v=\lim _{\delta t \rightarrow 0} \frac{r \cdot \delta s}{\delta t}=r \lim _{\delta t \rightarrow 0} \frac{\delta \theta}{\delta t}\)

But, \(\lim _{\delta t \rightarrow 0} \frac{\delta \theta}{\delta \mathrm{t}}=\omega\) = angular velocity

V= r. ω [for circular motion only]

i.e. (linear velocity) = (Radian) × (angular velocity)

In vector notation, \(\vec{v}=\vec{\omega} \times \vec{r}\) [in general]

The linear velocity of a particle performing circular motion is the vector product of its angular velocity and radius vector.

Relation Between Angular Acceleration And Linear Acceleration

For perfect circular motion, we know

v = ω r

on differentiating with respect to time

we get \(\frac{d v}{d t}=r \frac{d \omega}{d t}\)

a = r α

In vector form = \(\overrightarrow{\mathrm{a}}=\vec{\alpha} \times \overrightarrow{\mathrm{r}}\) ×(linear acc.) = (angular acc) × (radius)

Types Of Circular Motion

Uniform Circular Motion: The motion of a particle along the circumference of a circle with a constant speed is called uniform circular motion. Uniform circular motion is an accelerated motion. In the case of uniform circular motion :

Speed remains constant. v = constant

and v = ω r

angular velocity ω = constant

Motion will be periodic with time period = \(T=\frac{2 \pi}{\omega}=\frac{2 \pi r}{v}\)

Frequency Of Uniform Circular Motion: The number of revolutions performed per unit of time by the particle performing uniform circular motion is called the frequency (n)

∴ n = \(n=\frac{1}{T}=\frac{v}{2 \pi r}=\frac{\omega}{2 \pi}\)

S.I. unit of frequency is Hz.

As ω = constant, from ω = ω₀+ αt

angular acceleration α = 0

As a_t = αr, tangential acc. a_t= 0

As a_t = 0, a = \(a=\left(a_r^2+a_t^2\right)^{1 / 2}\) yields a = ar, i.e. acceleration is not zero but along radius towards the center and has magnitude

⇒ \(\mathrm{a}=\mathrm{a}_{\mathrm{r}}=\left(\mathrm{v}^2 / \mathrm{r}\right)=\mathrm{r} \omega^2\)

Speed and magnitude of acceleration are constant. but their directions are always changing so velocity and acceleration are not constant.

The direction of \(\overrightarrow{\mathrm{v}}\) is always along the tangent while that of \(\overrightarrow{\mathrm{a}_{\mathrm{r}}}\) along the radius \(\vec{v} \perp \vec{a}_r\)

If the moving body comes to rest, i.e. \(\vec{v} \rightarrow 0\), and if radial acceleration vanishes, the body will fly off along the tangent. So a tangential velocity and a radial acceleration (hence force) is a must for uniform circular motion.

As \(\vec{F}=\frac{m v^2}{r}\) ≠ 0, the body is not in equilibrium and the linear momentum of the particle moving on the circle is not conserved. However, as the force is control, i.e.,

⇒ \(\vec{\tau}=0\), so angular momentum is conserved, i.e.,

⇒ \(\overrightarrow{\mathrm{p}}\) ≠ constant but

⇒ \(\overrightarrow{\mathrm{L}}\) = constant

The work done by a centripetal force is always zero as it is perpendicular to velocity and hence displacement. By work-energy theorem as work done = change in kinetic energy ΔK = 0

So K (kinetic energy) remains constant

For example., Planets revolving around the sun, the motion of an electron around the nucleus in an atom

In one-dimensional motion, acceleration is always parallel to velocity and changes only the magnitude of the velocity vector.

In uniform circular motion, acceleration is always perpendicular to velocity and changes only the direction of the velocity vector.

In the more general case, like projectile motion, acceleration is neither parallel nor perpendicular to the figure that summarizes these three cases.

If a particle moving with uniform speed v on a circle of radius r suffers angular displacement θ in time Δt then change in its velocity.

⇒ \(\Delta \vec{v}=\Delta \vec{v}_2-\Delta \vec{v}_1\)

⇒ \(\vec{v}_1=\vec{v}_1 \hat{i}\)

⇒ \(\vec{v}_2=\vec{v}_2 \cos \theta \hat{i}+\vec{v}_2 \sin \theta \hat{j}\)

⇒ \(\Delta \vec{v}=\left(\vec{v}_2 \cos \theta-\vec{v}_1\right) \hat{i}+\vec{v}_2 \sin ^2 \hat{j}\)

⇒ \(|\Delta \vec{v}|=\sqrt{\left(\vec{v}_2 \cos \theta-\vec{v}_1\right)^2+\vec{v}_2 \sin ^2}\)

⇒ \(|\Delta \vec{v}|=\sqrt{2 v^2-2 v^2 \cos \theta}=\sqrt{2 v^2(1-\cos \theta)}=\sqrt{2 v^2\left(2 \sin ^2 \frac{\theta}{2}\right)}\)

⇒ \(v_1=v_2=v\)

⇒ \(|\Delta \vec{v}|=2 v \sin \frac{\theta}{2}\)

Question 1. A particle is moving in a circle of radius r centered at O with constant speed v. What is the change in velocity in moving from A to B? Given ∠AOB = 40º.
Answer:

⇒ \(|\Delta \vec{v}|=2 v \sin 40^{\circ} / 2=2 \mathrm{v} \sin 20^{\circ}\)

Non-Uniform Circular Motion:

A circular motion in which both the direction and magnitude of the velocity change is called nonuniform circular motion.

• A merry-go-round is spinning up from rest to full speed, or a ball whirling around in a vertical circle. The acceleration is neither parallel nor perpendicular to the velocity.
• We can resolve the acceleration vector into two components:

Radial Acceleration: ar perpendicular to the velocity ⇒ changes only the directions of velocity Acts just like the acceleration in a uniform circular motion.

⇒ \(a_c=\text { or } \quad a_r=\frac{v^2}{r}\)

Centripetal force: \(F_c=\frac{m v^2}{r}=m \omega^2 r\)

Tangential acceleration: a_r parallel to the velocity (since it is tangent to the path)

⇒ changes in the magnitude of the velocity act just like one-dimensional acceleration

⇒ \(a_t=\frac{d v}{d t}\)

Tangential acceleration : \(a_t=\frac{d v}{d t}\) where \(v=\frac{d s}{d t}\) and s = length of arc

Tangential acceleration: F_t = ma_t

The net acceleration vector is obtained by vector addition of these two components.

⇒ \(a=\sqrt{a_r^2+a_t^2}\)

In non-uniform circular motion :

speed \(|\vec{v}|\) ≠ constant angular velocity ω ≠ constant

i.e. speed ≠ constant i.e. angular velocity ≠ constant

In any instant

⇒ v = magnitude of the velocity of a particle

⇒ r = radius of circular path

⇒ ω = angular velocity of a particle

then, at that instant v = r ω

The net force on the particle

⇒ \(\vec{F}=\vec{F}_c+\vec{F}_t \Rightarrow F=\sqrt{F_c^2+F_t^2}\)

If θ is the angle made by F = Fc,

then tan θ = \(=\frac{F_t}{F_c} \Rightarrow \theta=\tan^{-1}\left[\frac{F_{\mathrm{t}}}{F_c}\right]\)

[Note angle between Fcand Ftis 90º] Angle between F and Ftis (90º – θ)

Net acceleration: \(a=\sqrt{a_c^2+a_1^2}=\frac{F_{\mathrm{net}}}{m}\)

The angle made by ‘a’ with ac, tan θ = \(\frac{a_t}{a_c}=\frac{F_t}{F_c}\)

Special Note:

In both uniform and non-uniform circular motion Fc is perpendicular to velocity.

So work done by centripetal force will be zero in both cases.

In uniform circular motion Ft= 0, as = at= 0, so work done will be zero by tangential force.

But in non-uniform circular motion Ft≠ 0, the work done by tangential force is non-zero.

Rate of work done by net force in non-uniform circular motion = rate of work done by tangential force

⇒ \(P=\frac{d W}{d t}=\vec{F}_t \cdot \vec{v}=\vec{F}_t \cdot \frac{d \vec{x}}{d t}\)

In a circle tangent and radius are always normal to each other, so

⇒ \(\vec{a}_{\mathrm{t}} \perp \vec{a}_{\mathrm{r}}\)

Net acceleration in case of circular motion \(a=a_r^2=a_t^2\)

Here it must be noted that at governs the magnitude of \(\vec{v}\) while ar its direction of motion so that

If ar = 0 and at = 0 a → 0 ⇒ motion is uniform translatory

If ar = 0 and at ≠ 0 a → at ⇒ motion is accelerated translatory

If ar ≠ 0 and at = 0 a → ar ⇒ motion is uniform circular

If ar ≠ 0 and at ≠ 0 a → \(a \rightarrow \sqrt{a_r^2+a_1^2}\) ⇒ motion is non-uniform circular.

Question 2. A road makes a 90º bend with a radius of 190 m. A car enters the bend moving at 20 m/s. Finding this too fast, the driver decelerates at 0.92 m/s². Determine the acceleration of the car when its speed rounding the bend has dropped to 15 m/s.
Answer:

Since it is rounding a curve, the car has a radial acceleration associated with its changing direction, in addition to the tangential deceleration that changes its speed. We are given that at = 0.92 m/s²; since the car is slowing down, the tangential acceleration is directed opposite the velocity.

The radial acceleration is \(a_r=\frac{v^2}{r}=\frac{(15 \mathrm{~m} / \mathrm{s})^2}{190 \mathrm{~m}}=1.2 \mathrm{~m} / \mathrm{s}^{21}\)

Magnitude of net acceleration,

⇒ \(a=\sqrt{a_r^2+a_t^2}=\left[(1.2 \mathrm{~m} / \mathrm{s})^2+(0.92 \mathrm{~m} / \mathrm{s})^2\right]^{1 / 2}=1.5 \mathrm{~m} / \mathrm{s}^2\)

and points at an angle \(\theta=\tan ^{-1}\left(\frac{\mathrm{a}_{\mathrm{r}}}{\mathrm{a}_{\mathrm{t}}}\right)=\tan ^{-1}\left(\frac{1.2 \mathrm{~m} / \mathrm{s}^2}{0.92 \mathrm{~m} / \mathrm{s}^2}\right)=53^{\circ}\)

relative to the tangent line to the circle.

Question 3. A particle is constrained to move in a circular path of radius r = 6m. Its velocity varies with time according to the relation v = 2t (m/s). Determine its

1. Centripetal acceleration,
2. Tangential acceleration,
3. Instantaneous acceleration at
   1. t = 0 sec. and
   2. t = 3 sec.

Answer:

At = 0,

v = 0, Thus a_r = 0

but \(\frac{d v}{d t}=2\) thus at = 2 m/s2 and a = \(\sqrt{a_t^2+a_r^2}=2 \mathrm{~m} / \mathrm{s}^2\)

At t = 3 sec. v = 6 m/s so \(a_r=\frac{v^2}{r}=\frac{(6)^2}{6}=6 \mathrm{~m} / \mathrm{s}^2\)

and \(a_t=\frac{d v}{d t}=2 \mathrm{~m} / \mathrm{s}^2\) Therefore, \(a=a=\sqrt{2^2+6^2}=\sqrt{40} \mathrm{~m} / \mathrm{s}^2\)

Question 4. The kinetic energy of a particle moving along a circle of radius r depends on the distance covered as K = As² where A is a constant. Find the force acting on the particle as a function of s.
Answer:

According to the given Question

⇒ \(\frac{1}{2} m v^2=A s^2 \text { or } v=s \sqrt{\frac{2 A}{m}}\) ……….(1)

So \(a_{\mathrm{r}}=\frac{\mathrm{v}^2}{\mathrm{r}}=\frac{2 \mathrm{A} \mathrm{s}^2}{\mathrm{mr}}\) ………….(2)

Furthermore as at = \(a_t=\frac{d v}{d t}=\frac{d v}{d s} \cdot \frac{d s}{d t}=v \frac{d v}{d s}\) …………(3)

from eqn. (1), ⇒ ……….. (4)

Substitute values from eqn(1) and eqn(4) in eqn(3)

⇒ \(a_t=\left[s \sqrt{\frac{2 A}{m}}\right]\left[\sqrt{\frac{2 A}{m}}\right]=\frac{2 A s}{m}\)

so \(a=\sqrt{a_r^2+a_t^2}=\sqrt{\left[\frac{2 A s^2}{m r}\right]^2+\left[\frac{2 A s}{m}\right]^2}\)

i.e. \(\mathrm{a}=\frac{2 \mathrm{As}}{\mathrm{m}} \sqrt{1+[\mathrm{s} / \mathrm{r}]^2}\)

so \(\mathrm{F}=\mathrm{ma}=2 \mathrm{As} \sqrt{1+[\mathrm{s} / \mathrm{r}]^2}\)

Question 5. A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration varies with time t as ac= k²rt², where k is a constant. Determine the power delivered to a particle by the forces acting on it.
Answer:

If v is instantaneous velocity, centripetal acceleration ac = \(a_c=\frac{v^2}{r} \Rightarrow=\frac{v^2}{r} k^2 r^2 \Rightarrow v=k r t\)

In circular motion work done by centripetal force is always zero and work is done only by tangential force.

∵ Tangent acceleration \(a_t=\frac{d v}{d t}=\frac{d}{d t}(k r t)=k r\)

∴ Tangential force Ft= mat= mkr

Power P = \(F_t v=(m k r)(k r t)=m k^2 r^2 t\)